 Myself, Satish Thalange, Assistant Professor, Department of Civil Engineering, Washington Institute of Technology, Salahpur. In today's session, we are going to discuss regarding the a solution of transportation problem by Vogel's approximation method. At the end of the session, the students will be able to determine the schedule of the transportation to minimize the transportation cost by using the Vogel's approximation method technique. Here, in the earlier lecture, we have seen what is a transportation problem. Now, we will see the particular transportation problem which was solved by the least cost method and the northwest corner method. Now, same problem, we will solve it with the Vogel's approximation method. This was the problem for the solving the transportation problem by VS method. Here, in this problem, before starting, we have to see the particular given problem is a balanced type or unbalanced type. Now, this is a balanced type because here, when we make the summation of the 5 plus 8 plus 7 plus 14, it is 34 and again, the summation of 7 plus 9 plus 18 is 34 which is the supply in the demand is equal. That is why it is a balanced transportation problem. Now, let us see the steps to be carried for the solution to solve the particular problem by the VM method. Here, the first step is to select the two smallest units cost from each row and to define the difference between it which will give the penalties. Similarly, we have to carry for the column also. After finding the penalties for each rows and columns, we have to select the maximum of it. It may be a row or it may be a column. After the selection of the row or the column, in that, we have to select the smallest transportation cost and we have to balance the supply and the demand. Let us see the problem and we will start to solve it. Here in the given problem, in the first step, the first row, there are the sources S1, S2, S3 and the designation D1, D2, D3 and D4. Here, we see that in the first row, the first smallest is 10 and the second smallest is 19. The difference between it is a 9. Here, I have written that the penalty 1 and the row is 9. Similarly, in the second row, the smallest is 30 and the second smallest is 40 and the difference between the 30 and 40 is 10, which I have written here. Similarly, for the source 3 also, same methodology we have to carry for the designation that is for the columns D1, D2, D3 and D4. Here, again the first penalty for the D1 is 21, second is 22 and for the D3 is 10 and the D4 is 10. Now, in this, the highest penalty we have to see in the whole rows and the columns, the highest is 22, that is, which is lying in the D2 column. In the D2 column, we have the transportation cost, that is, 30, 30 and 8. Out of these 30, 30 and 8, the smallest is 8. We have selected this cell and when we see selected this cell, it is lying in the source 3. Here, the designation D3 has having the demand of 8 and the source 3 is 18. The particular designation D2 is satisfied by 8 and as it has been satisfied, there will be a row reduction in the next step. The transportation cost for this step is 64. Again, as we see here, there is a column reduction D2 and the D1, D2, D3 and D4 are the columns. Here, again, we have to find out the second penalties. In the first row, the first penalty is 9. For the second row, source 3, it is 20 and for source 3, it is 20. Again, for D1, D3 and D4, out of this, all second penalties, the highest one is 21. Again, the same, it is lying in the D1 designation and the source S1. In this, we have selected S1 because out of this 40, 70 and 19, the smallest is 19, that is why we have selected this cell. Again, we have to make the demand and supply balancing and we will get that the particular transportation for the steps 2 is 95 because 19 into 5. 5 as it has been, the D1 is satisfied by the source S1. Similarly, for the step 3, we have to see. Again, we have to find out the penalties, the penalty 3. Here, there is a column reduction D1 as it has been satisfied. Again, in this, the 50 is the highest and it is lying in the source S3 and out of this source S3, 17, 20, the smallest is the 20, which is lying at D4. Again, we have to, the D4 is having the demand of 14 and the source available with the particular S3 is 10. The totally S3 has been supplied and the balancing has been done. As 10 has been supplied, 20 into 10, we are getting the total transportation cost that is of 200 for the step 3. Similarly, for the step 4, we have to carry the similar steps. Again, we have to find out the penalty 4. Out of this penalty 4, 10, 50, 40 and 20, this 40, 20. Again, I will repeat 10, 20, the 50 is highest. We have selected the cell that is having the transportation cost 10 because it is as this is 50 is lying in the designation D4 and in D4, 16, 10, the smallest is 10. Again, we have to balance the supply and the demand. Here, the transportation cost, whatever you are getting, it is 10 into 20. Sorry, 10 into 2, that is 20. Steps 4, step 5 is similar as here. As the particular source is being supplied from the source, S1 is totally being supplied. Here, there is a row reduction. Again, in this step, we have selected the cell 60 and out of this, because we have the chance, in these steps, we are finding the transportation cost that is 16 into 2, that is 120. Here, in the step 6, only the one cell has been left, that is 40. Here, the transportation cost, whatever you are getting, is particularly 280 because only one cell has been left. The demand is of 7 and the supply is 7 available. It has been satisfied. And the total transportation cost for the step 6 is 280 for the step 6. When we compare this particular total VM method with the other method, we will get, we will observe that the VM method is getting least transportation cost. But before that, this is the total transportation cost for the VM method. And the same problem, when we solve it by the particular least cost method, here we are getting 814. And same problem, if we solve it by the northwest corner method, we are getting 1050. This is a slide where we see that the particular VM method is having the least transportation cost as compared to the northwest method and the particular least cost method. Here, this is a VM method which is getting 779 transportation cost and by the least cost method, it is 840 and by the northwest corner, 1015. Okay, means the VM method is a very effective technique to solve the particular transportation problem, which is helpful for the scheduling as well as for the minimum transportation cost. Now, let's select the correct answer for the today's questions. These are the references for the today's session. Thank you.